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EXPLANATION OF NITROGEN IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) May 3, 2015 Nitrogen is a chemical element with symbol N and atomic number 7. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this image of Nitrogen including the following ground state electron configuration: 1s2.2s2.2px1.2py1.2pz1. According to the “Ionization energies of the elements-WIKIPEDIA” we observe that E1 = 14.53414 eV, E2 = 29.6013 eV, E3 = 47.44924 eV, E4 = 77.4735 eV, E5 = 97.89 eV, E6 = 552.0718 eV and E7 = 667.046 eV. EXPLANATION OF E1 = 14.53414 eV, E2 = 29.6013 eV, AND E3 = 47.44924 eV ''' In my paper of 2008 I showed that the one electron ( say 2pz1 ) produces deformations of the spherical shels1s2 and 2s2. Whereas the three electrons 2px1 , 2py1 , and 2pz1 of charge (- 3e) because of symmetry cannot produce deformations of the spherical shells. ( See the paper in “User Kaliambos”). For example for a perfect screening due to spherical shells we expect to find an effective ζ = 3, because +7e -2e -2e = +3e. For simplicity we start with the third ionization energy E3 = 47.44924 eV = - E(2pz1). Here in the absence of the electrons 2px1 and 2py1 one gets the E(2pz1) which is the binding energy of the one electron with n = 2 based on the Bohr model. In this case of n = 2 we must find the effective ζ1 > 3, because the shells 1s2 and 2s2 of the charge (-4e) screen the charge (+7e ) of the nucleus. Of course under a perfect screening we should write E (2pz1) = (-13.6) ζ2/22 = (-13.6)32/22 = -30.6 eV However the 2pz1 penetrates the 2s2 which leads to the great deformations not only of spherical shells 1s2 and 2s2 but also of 2pz1. Thus, writing E3 = 47.44924 eV = - E(2pz1) = - (- 13.6) ζ12/22 we get ζ12 = 13.955659 and ζ1 = 3.7357 > 3 . Then adding the second electron 2py1 one gets the binding energy E (2pz1 + 2py1) = - E3 - E2. Thus, for calculating the E2 = 29.6013 eV we may write E2 = 29.6013 eV = - E3 - E(2pz1 + 2py1) Here E(2pz1 + 2py1) = 2(-13.6)ζ22/22 which is the binding energy of the two separated electrons of 2pz1 and 2py1. Note that because of the parallel spin (S =1) we observe mutual electric and magnetic repulsions. Thus we have a mutual electromagnetic repulsion Fem = Fe + Fm which leads to the deformation of 2pz1 and 2py1. In other words the two electrons of 2pz1 and 2py1 try to be far away out of the shells 1s2 and 2s2. Since E2 = 29.6013 eV we may rewrite E(2pz1 + 2py1) = 2(-13.6)ζ22/22 = - E3 - E2 = - 47.44924 - 29.6013 = - 77.05 eV Therefore one gets ζ22 = 11.33 and ζ2 = 3.366 > 3 Then adding the third electron 2px1 one gets the binding energy of the three separated electrons which circulate outside the shells 1s2 and 2s2 : E(2px1 + 2py1 + 2pz1) = 3(-13.6)ζ32/22 = - E1 - E2 - E3 = - 91.58468 Therefore we get ζ32 = 9 and ζ3 = 3 Here ζ3 = 3 means that the three electrons of 2px1, 2py1, and 2pz1 make a spherical shell. Under this symmetry we observe a perfect screening of the spherical shells 1s2 and 2s2 which lead to ζ3 = 3. However in the absence of 1px1 the two electrons of 2py1 and 2pz1 break the symmetry and lead to ζ2 = 3.366 > 3 . Moreover in the absence of 2px1 and 2py1 the one electron of 2pz1 breaks more the symmetry and leads to ζ1 = 3.7357>3. That is, ζ1 > ζ2 > ζ3 . '''EXPLANATION OF E4 = 77.4735 eV AND E5 = 97.89 eV For simplicity we start with the E5 = -E( 2s1). According to the quantum mechanics the one electron (2s1) penetrates the 1s2 shell. Thus it leads to greater deformations of both 1s2 and 2s1 spherical shells giving an effective ζ > 5 because the charge (- 2e) of the two electrons of 1s2 screens the charge (+7e) of nucleus. Since n = 2 we may write E5 = 97.89 eV = -E(2s1) = - (-13.6)ζ2/22 Therefore one gets ζ2 = 28.7912 and ζ = 5.4 > 5 Then adding the second electron of opposite spin we get the binding energy for n = 2 given by my formula of 2008: E(2s2) = + (16.95)ζ - 4.1 / 22 = - E4 - E5 = - 145.33944 eV Then we may rewrite (27.2ζ2 - 16.95ζ + 4.1) /22 - 175.3637 = 0 Or 6.8ζ2 - 4.24ζ - 174.3637 = 0 Thus, solving for ζ we get ζ = 5.4 > 5 In other words we observe that the repulsions 1s2 - 2s1 and 1s2 - 2s2 give the same effective ζ = 5.4 > 5, because the two electrons (2s2) of opposite spin behave like one particle. Whereas in the case of the three electrons of 2p we observe a perfect screening, because the three electrons of 2px1, 2py1, and 2pz1 , interact from symmetrical positions. ' ' EXPLANATION OF Ε6 = 552.0718 eV AND E7 = 667.046 eV As in the case of the helium the ionization energy E7 = 667.046 eV = - E(1s1) is due to the one remaining electron of 1s1 with n = 1 given by applying the simple Bohr model for Z = 7 as E7 = - (-13.6057)Z2 /12 = - (-13.6057)72/12 = 666.68 eV. Surprisingly here we see that the value of 666.68 eV given by the Bohr model is smaller than the experimental value of 667.046 eV. It is due to the fact that after the ionizations the nuclear charge becomes much greater than the electron charge. Under this condition I suggest that the n=1 becomes n < 1. Therefore the above equation could be written as E7 = 667.046 eV = (13.6057)Z2 /12 = (13.6057)72 /n2 Then solving for n we get n = 0.999725 In the same way for calculating the E6 = 552. 0718 eV we must write my formula of 2008 as E6 = 552.0718 eV = - 667.046 - E(1s2 ) = - 667.046 - [ (-27.21)72 + (16.95)7 - 4.1] / n2 Then solving for n we get n = 0.999845. Here we see that 0.999845 > 0.999725 because the second electron increases the electron charge with respect to the nuclear charge. ' ' 'IN THE ABSENCE OF A DETAILED KNOWLEDGE ABOUT THE SPINNING ELECTRONS OF OPPOSITE SPIN THE VARIOUS WRONG THEORIES LEAD TO COMPLICATIONS ' It is of interest to note that the E(1s2) and the E(2s2) of spinning electrons of opposite spin with n = 1 and n = 2 respectively are given by applying my formula of my paper of 2008. However In the absence of a detailed knowledge about the electromagnetic force between the two spinning electrons of opposite spin physicists today using wrong theories cannot explain the ionization energies of nitrogen. For example under wrong theories based on qualitative approaches many physicists believe incorrectly that the second electron of the 1s2 shell is less tightly bound because it could be interpreted as a shielding effect; the other electron partly shields the second electron from the full charge of the nucleus. Another wrong way to view the energy is to say that the repulsion of the electrons contributes a positive potential energy which partially offsets the negative potential energy contributed by the attractive electric force of the nuclear charge. In atomic physics a two-electron atom is a quantum mechanical system consisting of one nucleus with a charge Ze and just two electrons. This is the first case of many-electron systems. The first few two-electron atoms are: Z =1 : H- hydrogen anion. Z = 2 : He helium atom. Z = 3 : Li+ lithium atom anion. Z = 4 : Be2+ beryllium ion. Z = 5 : B3+ boron. Prior to the development of quantum mechanics, an atom with many electrons was portrayed like the solar system, with the electrons representing the planets circulating about the nuclear “sun”. In the solar system, the gravitational interaction between planets is quite small compared with that between any planet and the very massive sun; interplanetary interactions can, therefore, be treated as small perturbations. However, in the helium atom with two electrons, the interaction energy between the two spinning electrons and between an electron and the nucleus are almost of the same magnitude, and a perturbation approach is inapplicable. In 1925 the two young Dutch physicists Uhlenbeck and Goudsmit discovered the electron spin according to which the peripheral velocity of a spinning electron is greater than the speed of light. Since this discovery invalidates Einstein’s relativity it met much opposition by physicists including Pauli. Under the influence of Einstein’s invalid relativity physicists believed that in nature cannot exist velocities faster than the speed of light.(See my FASTER THAN LIGHT). So great physicists like Pauli, Heisenberg, and Dirac abandoned the natural laws of electromagnetism in favor of wrong theories including qualitative approaches under an idea of symmetry properties between the two electrons of opposite spin which lead to many complications. Thus in the “Helium atom-Wikipedia” one reads: “Unlike for hydrogen a closed form solution to the Schrodinger equation for the helium atom has not been found. However various approximations such as the Hartree-Fock method ,can be used to estimate the ground state energy and wave function of atoms”. It is of interest to note that in 1993 in Olympia of Greece I presented at the international conference “Frontiers of fundamental physics” my paper “Impact of Maxwell’s equation of displacement current on electromagnetic laws and comparison of the Maxwellian waves with our model of dipolic particles " The conference was organized by the natural philosophers M. Barone and F. Selleri who awarded me an award including a disc of the atomic philospher Democritus, because . in that paper I showed that LAWS AND EXPERIMENTS INVALIDATE FIELDS AND RELATIVITY . At the same period I tried to find not only the nuclear force and structure but also the coupling of two electrons under the application of the abandoned electromagnetic laws. For example in the photoelectric effect the absorption of light contributes not only to the increase of the electron energy but also to the increase of the electron mass because the particles of light have mass m = hν/c2 .( See my DISCOVERY OF PHOTON MASS ). Under this condition of photon mass absorption the electron cannot move faster than light. (υ2/Mo2 = c2 /(c2-υ2) Note that differentiation of this equation led to my discovery of PHOTON- MATTER INTERACTION: hν/m = ΔΕ/ΔΜ = c2 However the electron spin which gives a peripheral velocity greater than the speed of light cannot be affected by the photon absorption. Thus after 10 years I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism" (2003), in which I showed not only my DISCOVERY OF NUCLEAR FORCE AND STRUCTURE but also that the peripheral velocity (u >> c) of two spinning electrons with opposite spin gives an attractive magnetic force (Fm) stronger than the electric repulsion (Fe) when the two electrons of mass m and charge (-e) are at a very short separation r < 578.8 /1015 m. Because of the antiparallel spin along the radial direction the interaction of the electron charges gives an electromagnetic force Fem = Fe - Fm . Therefore in my research the integration for calculating the mutual Fem led to the following relation: Fem = Fe - Fm = Ke2/r2 - (Ke2/r4)(9h2/16π2m2c2) Of course for Fe = Fm one gets the equilibrium separation ro = 3h/4πmc = 578.8/1015 m. That is, for r < 578.8/1015 m the two electrons of opposite spin exert an attractive electromagnetic force, because the attractive Fm is stronger than the repulsive Fe . Here Fm is a spin-dependent force of short range. As a consequence this situation provides the physical basis for understanding the pairing of two electrons described qualitatively by the Pauli principle, which cannot be applied in the simplest case of the deuteron in nuclear physics, because the binding energy between the two spinning nucleons occurs when the spin is not opposite (S=0) but parallel (S=1). According to the experiments in the case of two electrons with antiparallel spin the presence of a very strong external magnetic force gives parallel spin (S=1) with electric and magnetic repulsions given by Fem = Fe + Fm So, according to the well-established laws of electromagnetism after a detailed analysis of paired electrons in two-electron atoms I concluded that at r < 578.8/1015 m a motional EMF produces vibrations of paired electrons. Unfortunately today many physicists in the absence of a detailed knowledge believe that the two electrons of two-electron atoms under the Coulomb repulsion between the electrons move not together as one particle but as separated particles possessing the two opposite points of the diameter of the orbit around the nucleus. In fact, the two electrons of opposite spin behave like one particle circulating about the nucleus under the rules of quantum mechanics forming two-electron orbitals in helium, beryllium etc. In my paper of 2008, I showed that the positive vibration energy (Ev) described in eV depends on the Ze charge of nucleus as Ev = 16.95Z - 4.1 Of course in the absence of such a vibration energy Ev it is well-known that the ground state energy E described in eV for two orbiting electrons could be given by the Bohr model as E = (-27.2) Z2. So the combination of the energies of the Bohr model and the vibration energies due to the opposite spin of two electrons led to my discovery of the ground state energy of two-electron atoms given by E = (-27.2) Z2 + (16.95 )Z - 4.1 For example the laboratory measurement of the ionization energy of H- yields an energy of the ground state E = - 14.35 eV. In this case since Z = 1 we get E -27.2 + 16.95 - 4.1 = -14.35 eV. In the same way writing for the helium Z = 2 we get E = - 108.8 + 32.9 - 4.1 = -79.0 eV which is equal to the laboratory measurement. In the same way we can calculate the ground state energies for the Z = 3 : Li+ ion , Z = 4 : Be2+ beryllium ion, and Z = 5 : B3+ boron. The discovery of this simple formula based on the well-established laws of electromagnetism was the first fundamental equation for understanding the energies of many-electron atoms, while various theories based on qualitative symmetry properties lead to complications. Category:Fundamental physics concepts